Dayo Ogundipe edited begin_itemize_item_textbf_Show__.tex  over 8 years ago

Commit id: 80e8ca91b1efdd82bce2493d2b7d5151f74db8bf

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\item \textbf{Show that $\log_{10}(2)$ is not a rational number.}  \[ \log_{10}(2) = \frac{p}{q}\] where p,q are integers  \[ 10^{\frac{p}{q}} = 2\]  \[ 10^{p} - 10^{q} 10^{p-q}}  = 2 \] $ \nexists$ two integers for p,q.  \\Also  \[ (10^{\frac{p}{q}})^q = 2^q\] 

$m = a_{1}^{\mu_{1}} , a_{2}^{\mu_{2}}, a_{3}^{\mu_{3}}, a_{i}^{\mu_{i}}$ where $\mu_{i}$ are integers $\ge 1$ and $a_{i}$ are primes.  \\ $n = b_{1}^{v_{1}} , b_{2}^{v_{2}}, b_{3}^{v_{3}}, b_{i}^{v_{i}}$ where $v_{i}$ are integers $\ge 1$ and $b_{i}$ are primes.  \[ m^p - m^q\] m^{p-q} = n\]  \end{itemize}